How can I prove that all the a-cuts of any fuzzy set A defined on $R^n$ are convex?
We can assume without loss of generality that
\begin{equation}
\min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha.
\end{equation}
$\implies$

a-cut of fuzzy set $A$ is on $R^n$ is convex. A-cut can be defined as
\begin{equation}
A = \{x \in R^n| \mu_A(x) \geq \alpha\}
\end{equation}
If we take two elements $r$ and $s$, by the definition of convex set, number $\lambda r + (1 + \lambda)s$ is also an element of that set. Since it's an element of that set that means
\begin{equation}
\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha
\end{equation}

$\impliedby$

$\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$.

We know from $\min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha$ that $\mu_A(s) &gt; \alpha$. We have an affine combination $\lambda r + (1 + \lambda)s$ for which also $\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$ so we know that all numbers $\lambda r + (1 + \lambda)s$ satisfy inequality $\mu_A(\cdot) \geq \alpha$ (belong to the same set as $r$ and $s$) which means this is a convex set again by the definition of a convex set.